What is the rank (minimal number of group generators) of $SL(n,\mathbb{F})$ in the situation when $SL(n,\mathbb{F})$ is not perfect (i.e. when $SL(n,\mathbb{F})$ is different from $SL(2,\mathbb{F}_2)$ and $SL(2,\mathbb{F}_3)$)? If the general formula is not known, are there $n$ and $F$ which make $rank(SL(n,\mathbb{F}))>k$ for a fixed $k$ (or at least for $k=2$)?
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2$\begingroup$ You mean when SL(n,F) is perfect perhaps? $\endgroup$– Geoff RobinsonCommented Apr 10, 2015 at 18:09
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3 Answers
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L.E. Dickson proved that when $\mathbb{F}$ is a finite field of odd characteristic $p$, then ${\rm SL}(2,\mathbb{F}) = \langle \left(\begin{array}{clcr} 1&1\\0&1\end{array} \right), \left(\begin{array}{clcr} 1&0\\\lambda &1\end{array} \right) \rangle$ when $\lambda$ is a generator of the multiplicative group of $\mathbb{F}$, except when $|\mathbb{F}| = 9.$ Also, ( a very special case of a much more general result of) R. Steinberg proved that ${\rm PSL}(n,\mathbb{F})$ is generated by two elements whenever $\mathbb{F}$ is a finite field and ${\rm PSL}(n,\mathbb{F})$ is simple. Since the center of ${\rm SL}(n,\mathbb{F})$ is contained in its Frattini subgroup in that case, we know that ${\rm SL}(n,\mathbb{F})$ is still generated by $2$ elements.
answered Apr 10, 2015 at 18:40
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$\begingroup$ Are the two elements explicit, as they are in the case of $\mathbb{F}_p?$ (as per my answer? $\endgroup$ Commented Apr 10, 2015 at 18:41
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$\begingroup$ In Steinberg's case they are explicitly described. One is the longest element of the Weyl group, for example. $\endgroup$ Commented Apr 10, 2015 at 18:43
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$\begingroup$ As Geoff says, Steinberg proved this statement in a much more general context for groups of Lie type, following a long history of developments which he refers to. His paper is here: cms.math.ca/10.4153/CJM-1962-018-0 $\endgroup$ Commented Apr 10, 2015 at 22:28
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For any finite field $F$ and $n \ge 2$, ${\rm SL}(n,F)$ is generated by $A$ and $B$, where $A$ is the diagonal matrix with entries $(\omega,\omega^{-1},1,1,\ldots,1)$, for a primitive field element $\omega$ and $$B = \left(\begin{array}{ccccccc}-1&0&0&\ldots&0&0&1\\ -1&0&0&\ldots&0&0&0\\ 0&-1&0&\ldots&0&0&0\\ 0&0&-1&\ldots&0&0&0\\ &&&\ldots&&&&\\ 0&0&0&\ldots&0&-1&0\\ \end{array}\right).$$
answered Apr 10, 2015 at 20:17
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Since $SL(n, \mathbb{Z})$ is generated by two matrices (a result of Hua and Reiner from 1948), the same is true for $SL(n, \mathbb{Z}/p)$ for any prime $p.$
answered Apr 10, 2015 at 17:54
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$\begingroup$ Thank you, but is something known about the case of general char p fields, i.e. when F has p^k elements? Because I am trying to understand if the rank in certain situations can be bigger than 2. $\endgroup$– user70387Commented Apr 10, 2015 at 18:16
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$\begingroup$ ${\rm SL}(n,F)$ can be generated by two elements for all finite fields $F$. $\endgroup$ Commented Apr 10, 2015 at 20:01